Law of Sines Word Problems: Surveying, Navigation and Heights

A surveyor on a riverbank sights a tree on the opposite shore. She measures a 50-metre baseline along her bank, then reads angles of 75° and 80° from each end to the tree, and this is one of those law of sines word problems that require setting up the triangle from the given data. She does not wade across. She reaches for the law of sines. Set up the same applied problems you will meet in class and in the field: sketch the triangle, label correctly, and pick the law that fits the given data.

Draw the triangle. Label every known angle and side. Identify what you are asked to find. Then decide which law fits. The law of sines works when you know two angles and a side (AAS or ASA) or two sides and a non-included angle (SSA). For SSA, stop and check, this is the ambiguous case, and you may get zero, one, or two valid triangles. Stewart/Redlin/Watson's Precalculus gives the rule: given side a, side b, and angle A opposite a, zero solutions exist when a < b sin A; one solution when a = b sin A (right triangle) or a ≥ b; two solutions when b sin A < a < b. Verify the angle mode. A theodolite measures in degrees, minutes, seconds. Your calculator must be in degree mode, not radians. A single mis-mode step makes every subsequent number wrong.

Why the Sketch Matters

A good sketch fixes the known baseline opposite the measured angles. Label the baseline side c, the other sides a and b, and the opposite angles A, B, C. In a triangulation problem, the baseline is the only known distance, everything else comes from the law of sines. The sketch prevents you from applying the law of sines to an SAS situation (two sides and the included angle), which needs the law of cosines instead.

Surveyors use triangulation when direct measurement is impossible, across a river, a gorge, or a stretch of swamp. Ghilani and Wolf's Elementary Surveying (15th ed., 2018) describes the standard method: measure a baseline to an accuracy of 1:10,000 or better, then measure horizontal angles to a target with a theodolite to the nearest second of arc. Here is a worked example that appears in surveying field manuals.

Example 1: River Crossing. A surveyor measures baseline AB = 250.00 m along one bank. From station A, the angle to a tree C on the opposite bank is 48°12′30″. From station B, the angle to the same tree is 72°08′45″. Find the distance AC from station A to the tree.

Sketch: Triangle ABC with baseline AB = 250.00 m, angle A = 48°12′30″, angle B = 72°08′45″. Angle C = 180° − (48°12′30″ + 72°08′45″) = 59°38′45″. Use the law of sines: AC / sin B = AB / sin C. So AC = (AB × sin B) / sin C = (250.00 × sin 72°08′45″) / sin 59°38′45″.8616. AC ≈ (250.00 × 0.9512) / 0.8616 ≈ 276.0 m. The surveyor records AC = 276.0 m to the nearest 0.1 m.For a first-order survey, Ghilani and Wolf require closure within ± 10″ in angle sum; keep six decimal places throughout.

Navigation problems use bearings, directions measured clockwise from north, written as NθE or SθW, where θ is an acute angle from the north-south axis. Stewart/Redlin/Watson's Precalculus uses this convention. A bearing of N30°E means 30° east of due north. A bearing of S45°W means 45° west of due south. Convert the bearing to an interior triangle angle before applying the law of sines. Law of sines bearing problems appear in every precalculus text and in navigation training for backup positioning when GPS is unavailable.

Example 2: Ship Position from Two Lighthouses. A sailor sights lighthouse A at a bearing of N30°E and lighthouse B at a bearing of N75°E. The chart shows the distance between the lighthouses is 8.0 nautical miles. Find the ship's distance to each lighthouse.

Sketch: Place the ship at vertex C. Light A at vertex A, light B at vertex B. The bearing from ship to A is N30°E, so the angle at C between north and CA is 30°. The bearing to B is N75°E, so the angle between north and CB is 75°. The interior angle at C is the difference: 75° − 30° = 45°. The baseline AB = 8.0 nautical miles. You need one more angle. The problem gives bearings from the ship, not angles at the lighthouses. To find angle A, draw a line north from A. The bearing from A to the ship is the back bearing, the opposite direction. The back bearing of N30°E is S30°W, which makes an interior angle at A of 30° (south of west). Similarly, the back bearing of N75°E is S75°W, giving an interior angle at B of 75°. But interior angles must sum to 180°. Check: C = 45°, A = 30°, B = 75°? That sums to 150°, not 180°. The error: interior angles at A and B are not simply the bearing angles. Instead, use the fact that the straight line AB runs between the lighthouses; the angles at A and B are supplementary to the bearings. A full solution requires drawing the north-south lines at each point. The correct interior angles are: A = 30° (angle between AB and the north line at A), B = 15° (angle between BA and the north line at B), C = 135° (the bearing difference is 45°, but the interior angle is 180° − 45° = 135°). Sum = 180°. Now apply law of sines: AC / sin B = AB / sin C. AC = (8.0 × sin 15°) / sin 135° ≈ (8.0 × 0.2588) / 0.7071 ≈ 2.93 nautical miles. BC = (8.0 × sin 30°) / sin 135° ≈ (8.0 × 0.5) / 0.7071 ≈ 5.66 nautical miles. The ship is 2.9 NM from light A, 5.7 NM from light B.

Common Failure: Ignoring the Back Bearing

Students often take the given bearing as the interior angle without converting. That yields a triangle sum well short of 180° and a nonsense result. The fix: always draw north arrows at every point and work out interior angles by taking differences or using supplementary relationships.

Finding the height of a cliff or a building from two angles of elevation is a classic triangulation problem. You measure the angle to the top from two points at known horizontal distance apart. The height is the side opposite the vertical angle in a right triangle, but the base of the triangle is not horizontal ground, the two points are at the same level, and the vertical line from the top meets the baseline at an unknown horizontal distance. Use the law of sines on the oblique triangle formed by the two observation points and the top.

Example 3: Cliff Height. From point A, 100 m from the base of a cliff, the angle of elevation to the top is 35°. From point B, 200 m from the base (farther away), the angle of elevation is 20°. Find the height of the cliff.

Sketch: Points A and B are 100 m apart horizontally (difference of distances to base). The top of the cliff is point C. Triangle ABC: side AB = 100 m, angle at A = 35°, angle at B = 180° − 20° = 160°? No, the angles are measured from horizontal, not from the side of the triangle. The interior angle at A between the line AB and the line AC is 90° − 35° = 55° (since AC makes a 35° angle above horizontal, AB is horizontal). The interior angle at B between BA and BC is 90° − 20° = 70° (BC makes a 20° angle above horizontal, BA is horizontal in the opposite direction). Angle C = 180° − (55° + 70°) = 55°. Now apply law of sines: AC / sin B = AB / sin C. AC = (100 × sin 70°) / sin 55° ≈ (100 × 0.9397) / 0.8192 ≈ 114.7 m. The height h = AC × sin 35° = 114.7 × 0.5736 ≈ 65.8 m. The cliff is about 66 m high.

Why This Works

The law of sines handles the oblique triangle between the observer and the top. The vertical side is then found by dropping a perpendicular from the top to the baseline, a right-triangle step that uses the sine of the elevation angle. The method works for any two points that are level and whose distance apart you know.

In triangulation, a small error in an angle produces a large error in the computed distance. Ghilani and Wolf give the rule: a 1″ error in an angle at a distance of 1 km displaces the point by about 5 mm. That is 5 mm per km per arcsecond. A 0.1° error (360″) at 500 m shifts the computed point by 360 × 0.5 × 5 mm = 900 mm, or nearly a metre. Surveyors measure angles to the nearest second with a theodolite to keep error under 1:10,000. Navigators, using a hand compass to the nearest 0.5°, accept a displacement of tens of metres over a 10 km baseline. The law of sines magnifies angle errors because the side is proportional to sin(angle): near 90°, a small change in angle barely changes the sine; near 0° or 180°, a tiny angular change doubles the side error. In classroom word problems, the numbers come out clean because textbook authors pick neat angles. In the field, you must carry full precision through every step and round only at the final answer.

Example 4: Error Propagation. In Example 1, suppose the surveyor misreads angle A as 48°12′30″ instead of the true 48°13′00″, an error of 30″ (0.0083°). The computed AC changes from 276.1 m to 276.0 m, a shift of 0.1 m over 250 m baseline. That is a relative error of 0.04%, acceptable for third-order work. But if the baseline were 2 km and the angles were near 10°, the same 30″ error would shift the computed distance by several metres. The lesson: never round intermediate angles or sides. Store them in full calculator precision.